Integrand size = 42, antiderivative size = 28 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx=-\frac {3}{2}+e^x+x+\frac {1}{4} \left (\frac {3}{x}+\left (-e^x+x\right ) \log (x)\right ) \] Output:
x-3/2+exp(x)+1/4*(x-exp(x))*ln(x)+3/4/x
Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx=\frac {1}{4} \left (4 e^x+\frac {3}{x}+4 x-e^x \log (x)+x \log (x)\right ) \] Input:
Integrate[(-3 + 5*x^2 + E^x*(-x + 4*x^2) + (x^2 - E^x*x^2)*Log[x])/(4*x^2) ,x]
Output:
(4*E^x + 3/x + 4*x - E^x*Log[x] + x*Log[x])/4
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {27, 25, 2010, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {5 x^2+e^x \left (4 x^2-x\right )+\left (x^2-e^x x^2\right ) \log (x)-3}{4 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {-5 x^2+e^x \left (x-4 x^2\right )-\left (x^2-e^x x^2\right ) \log (x)+3}{x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {-5 x^2+e^x \left (x-4 x^2\right )-\left (x^2-e^x x^2\right ) \log (x)+3}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{4} \int \left (\frac {e^x (\log (x) x-4 x+1)}{x}+\frac {-\log (x) x^2-5 x^2+3}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (4 x+\frac {3}{x}+x \log (x)+\frac {e^x (4 x-x \log (x))}{x}\right )\) |
Input:
Int[(-3 + 5*x^2 + E^x*(-x + 4*x^2) + (x^2 - E^x*x^2)*Log[x])/(4*x^2),x]
Output:
(3/x + 4*x + x*Log[x] + (E^x*(4*x - x*Log[x]))/x)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.56 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
default | \(-\frac {{\mathrm e}^{x} \ln \left (x \right )}{4}+{\mathrm e}^{x}+x +\frac {3}{4 x}+\frac {x \ln \left (x \right )}{4}\) | \(21\) |
parts | \(-\frac {{\mathrm e}^{x} \ln \left (x \right )}{4}+{\mathrm e}^{x}+x +\frac {3}{4 x}+\frac {x \ln \left (x \right )}{4}\) | \(21\) |
norman | \(\frac {\frac {3}{4}+x^{2}+{\mathrm e}^{x} x +\frac {x^{2} \ln \left (x \right )}{4}-\frac {x \,{\mathrm e}^{x} \ln \left (x \right )}{4}}{x}\) | \(28\) |
risch | \(\frac {\left (x -{\mathrm e}^{x}\right ) \ln \left (x \right )}{4}+\frac {4 x^{2}+4 \,{\mathrm e}^{x} x +3}{4 x}\) | \(29\) |
parallelrisch | \(-\frac {x \,{\mathrm e}^{x} \ln \left (x \right )-x^{2} \ln \left (x \right )-4 \,{\mathrm e}^{x} x -4 x^{2}-3}{4 x}\) | \(31\) |
orering | \(\frac {\left (-{\mathrm e}^{x} x^{2}+x^{2}\right ) \ln \left (x \right )+\left (4 x^{2}-x \right ) {\mathrm e}^{x}+5 x^{2}-3}{4 x}-\frac {\left (x^{8}+x^{7}-9 x^{6}+30 x^{5}-4 x^{4}-4 x^{3}-178 x^{2}+108 x -36\right ) \left (\frac {\left (-{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 x \right ) \ln \left (x \right )+\frac {-{\mathrm e}^{x} x^{2}+x^{2}}{x}+\left (8 x -1\right ) {\mathrm e}^{x}+\left (4 x^{2}-x \right ) {\mathrm e}^{x}+10 x}{4 x^{2}}-\frac {\left (-{\mathrm e}^{x} x^{2}+x^{2}\right ) \ln \left (x \right )+\left (4 x^{2}-x \right ) {\mathrm e}^{x}+5 x^{2}-3}{2 x^{3}}\right )}{x^{6}-x^{5}+9 x^{4}+16 x^{3}+16 x^{2}-36 x +36}+\frac {2 x \left (x^{7}-7 x^{5}+13 x^{4}-2 x^{3}+10 x^{2}-66 x +18\right ) \left (\frac {\left (-{\mathrm e}^{x} x^{2}-4 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+2\right ) \ln \left (x \right )+\frac {-2 \,{\mathrm e}^{x} x^{2}-4 \,{\mathrm e}^{x} x +4 x}{x}-\frac {-{\mathrm e}^{x} x^{2}+x^{2}}{x^{2}}+8 \,{\mathrm e}^{x}+2 \left (8 x -1\right ) {\mathrm e}^{x}+\left (4 x^{2}-x \right ) {\mathrm e}^{x}+10}{4 x^{2}}-\frac {\left (-{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 x \right ) \ln \left (x \right )+\frac {-{\mathrm e}^{x} x^{2}+x^{2}}{x}+\left (8 x -1\right ) {\mathrm e}^{x}+\left (4 x^{2}-x \right ) {\mathrm e}^{x}+10 x}{x^{3}}+\frac {\frac {3 \left (-{\mathrm e}^{x} x^{2}+x^{2}\right ) \ln \left (x \right )}{2}+\frac {3 \left (4 x^{2}-x \right ) {\mathrm e}^{x}}{2}+\frac {15 x^{2}}{2}-\frac {9}{2}}{x^{4}}\right )}{x^{6}-x^{5}+9 x^{4}+16 x^{3}+16 x^{2}-36 x +36}-\frac {\left (x^{6}-7 x^{4}+6 x^{3}+7 x^{2}+24 x -6\right ) x^{2} \left (\frac {\left (-{\mathrm e}^{x} x^{2}-6 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}\right ) \ln \left (x \right )+\frac {-3 \,{\mathrm e}^{x} x^{2}-12 \,{\mathrm e}^{x} x -6 \,{\mathrm e}^{x}+6}{x}-\frac {3 \left (-{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 x \right )}{x^{2}}+\frac {-2 \,{\mathrm e}^{x} x^{2}+2 x^{2}}{x^{3}}+24 \,{\mathrm e}^{x}+3 \left (8 x -1\right ) {\mathrm e}^{x}+\left (4 x^{2}-x \right ) {\mathrm e}^{x}}{4 x^{2}}-\frac {3 \left (\left (-{\mathrm e}^{x} x^{2}-4 \,{\mathrm e}^{x} x -2 \,{\mathrm e}^{x}+2\right ) \ln \left (x \right )+\frac {-2 \,{\mathrm e}^{x} x^{2}-4 \,{\mathrm e}^{x} x +4 x}{x}-\frac {-{\mathrm e}^{x} x^{2}+x^{2}}{x^{2}}+8 \,{\mathrm e}^{x}+2 \left (8 x -1\right ) {\mathrm e}^{x}+\left (4 x^{2}-x \right ) {\mathrm e}^{x}+10\right )}{2 x^{3}}+\frac {\frac {9 \left (-{\mathrm e}^{x} x^{2}-2 \,{\mathrm e}^{x} x +2 x \right ) \ln \left (x \right )}{2}+\frac {9 \left (-{\mathrm e}^{x} x^{2}+x^{2}\right )}{2 x}+\frac {9 \left (8 x -1\right ) {\mathrm e}^{x}}{2}+\frac {9 \left (4 x^{2}-x \right ) {\mathrm e}^{x}}{2}+45 x}{x^{4}}-\frac {6 \left (\left (-{\mathrm e}^{x} x^{2}+x^{2}\right ) \ln \left (x \right )+\left (4 x^{2}-x \right ) {\mathrm e}^{x}+5 x^{2}-3\right )}{x^{5}}\right )}{x^{6}-x^{5}+9 x^{4}+16 x^{3}+16 x^{2}-36 x +36}\) | \(824\) |
Input:
int(1/4*((-exp(x)*x^2+x^2)*ln(x)+(4*x^2-x)*exp(x)+5*x^2-3)/x^2,x,method=_R ETURNVERBOSE)
Output:
-1/4*exp(x)*ln(x)+exp(x)+x+3/4/x+1/4*x*ln(x)
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx=\frac {4 \, x^{2} + 4 \, x e^{x} + {\left (x^{2} - x e^{x}\right )} \log \left (x\right ) + 3}{4 \, x} \] Input:
integrate(1/4*((-exp(x)*x^2+x^2)*log(x)+(4*x^2-x)*exp(x)+5*x^2-3)/x^2,x, a lgorithm="fricas")
Output:
1/4*(4*x^2 + 4*x*e^x + (x^2 - x*e^x)*log(x) + 3)/x
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx=\frac {x \log {\left (x \right )}}{4} + x + \frac {\left (4 - \log {\left (x \right )}\right ) e^{x}}{4} + \frac {3}{4 x} \] Input:
integrate(1/4*((-exp(x)*x**2+x**2)*ln(x)+(4*x**2-x)*exp(x)+5*x**2-3)/x**2, x)
Output:
x*log(x)/4 + x + (4 - log(x))*exp(x)/4 + 3/(4*x)
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx=\frac {1}{4} \, x \log \left (x\right ) - \frac {1}{4} \, e^{x} \log \left (x\right ) + x + \frac {3}{4 \, x} + e^{x} \] Input:
integrate(1/4*((-exp(x)*x^2+x^2)*log(x)+(4*x^2-x)*exp(x)+5*x^2-3)/x^2,x, a lgorithm="maxima")
Output:
1/4*x*log(x) - 1/4*e^x*log(x) + x + 3/4/x + e^x
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx=\frac {x^{2} \log \left (x\right ) - x e^{x} \log \left (x\right ) + 4 \, x^{2} + 4 \, x e^{x} + 3}{4 \, x} \] Input:
integrate(1/4*((-exp(x)*x^2+x^2)*log(x)+(4*x^2-x)*exp(x)+5*x^2-3)/x^2,x, a lgorithm="giac")
Output:
1/4*(x^2*log(x) - x*e^x*log(x) + 4*x^2 + 4*x*e^x + 3)/x
Time = 4.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx={\mathrm {e}}^x-\frac {{\mathrm {e}}^x\,\ln \left (x\right )}{4}+x\,\left (\frac {\ln \left (x\right )}{4}+1\right )+\frac {3}{4\,x} \] Input:
int(-((exp(x)*(x - 4*x^2))/4 - (5*x^2)/4 + (log(x)*(x^2*exp(x) - x^2))/4 + 3/4)/x^2,x)
Output:
exp(x) - (exp(x)*log(x))/4 + x*(log(x)/4 + 1) + 3/(4*x)
Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {-3+5 x^2+e^x \left (-x+4 x^2\right )+\left (x^2-e^x x^2\right ) \log (x)}{4 x^2} \, dx=\frac {-e^{x} \mathrm {log}\left (x \right ) x +4 e^{x} x +\mathrm {log}\left (x \right ) x^{2}+4 x^{2}+3}{4 x} \] Input:
int(1/4*((-exp(x)*x^2+x^2)*log(x)+(4*x^2-x)*exp(x)+5*x^2-3)/x^2,x)
Output:
( - e**x*log(x)*x + 4*e**x*x + log(x)*x**2 + 4*x**2 + 3)/(4*x)